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          算法练习题27---蓝桥杯2020省赛“装饰珠”
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        <h1 id="前言"><a href="#前言" class="headerlink" title="前言"></a>前言</h1><p>蓝桥杯2020省赛，编程题（C++）</p>
<span id="more"></span>

<h1 id="一、题目描述"><a href="#一、题目描述" class="headerlink" title="一、题目描述"></a>一、题目描述</h1><p>在怪物猎人这一款游戏中，玩家可以通过给装备镶嵌不同的装饰珠来获取 相应的技能，以提升自己的战斗能力。</p>
<p>已知猎人身上一共有 6 件装备，每件装备可能有若干个装饰孔，每个装饰孔有各自的等级，可以镶嵌一颗小于等于自身等级的装饰珠 (也可以选择不镶嵌)。</p>
<p>装饰珠有 M 种，编号 1 至 M，分别对应 M 种技能，第 i 种装饰珠的等级为 iLi，只能镶嵌在等级大于等于Li 的装饰孔中。</p>
<p>对第 i 种技能来说，当装备相应技能的装饰珠数量达到 Ki 个时，会产生 Wi(Ki) 的价值。镶嵌同类技能的数量越多，产生的价值越大，即 Wi(Ki−1)&lt;Wi(Ki)。但每个技能都有上限 Pi(1≤Pi≤7)，当装备的珠子数量超过 Pi 时，只会产生Wi(Pi) 的价值。</p>
<p>对于给定的装备和装饰珠数据，求解如何镶嵌装饰珠，使得 6 件装备能得到的总价值达到最大。</p>
<h3 id="输入描述"><a href="#输入描述" class="headerlink" title="输入描述"></a>输入描述</h3><p>输入的第 1 至 6 行，包含 6 件装备的描述。其中第 i 行的第一个整数 Ni 表示第 i 件装备的装饰孔数量。后面紧接着 Ni 个整数，分别表示该装备上每个装饰孔的等级 L (1≤L≤4)。</p>
<p>第 7 行包含一个正整数 M，表示装饰珠 (技能) 种类数量。</p>
<p>第 8 至 M + 7 行，每行描述一种装饰珠 (技能) 的情况。每行的前两个整数 Lj (1≤Lj≤4) 和 Pj (1≤Pi≤7) 分别表示第 j种装饰珠的等级和上限。接下来 Pj 个整数，其中第 k 个数表示该装备中装饰珠数量为 k 时的价值 Wj(k)。</p>
<p>其中，1≤Ni≤50,1≤M≤10^4,1≤Wj(k)≤10^4。</p>
<h3 id="输出描述"><a href="#输出描述" class="headerlink" title="输出描述"></a>输出描述</h3><p>输出一行包含一个整数，表示能够得到的最大价值。</p>
<h3 id="输入输出样例"><a href="#输入输出样例" class="headerlink" title="输入输出样例"></a>输入输出样例</h3><h4 id="示例"><a href="#示例" class="headerlink" title="示例"></a>示例</h4><blockquote>
<p>输入</p>
</blockquote>
<figure class="highlight plaintext"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br></pre></td><td class="code"><pre><span class="line">1 1</span><br><span class="line">2 1 2</span><br><span class="line">1 1</span><br><span class="line">2 2 2</span><br><span class="line">1 1</span><br><span class="line">1 3</span><br><span class="line">3</span><br><span class="line">1 5 1 2 3 5 8</span><br><span class="line">2 4 2 4 8 15</span><br><span class="line">3 2 5 10</span><br></pre></td></tr></table></figure>

<blockquote>
<p>输出</p>
</blockquote>
<figure class="highlight plaintext"><table><tr><td class="gutter"><pre><span class="line">1</span><br></pre></td><td class="code"><pre><span class="line">20</span><br></pre></td></tr></table></figure>

<blockquote>
<p>样例说明</p>
</blockquote>
<p>按照如下方式镶嵌珠子得到最大价值 20，括号内表示镶嵌的装饰珠的种类编号：</p>
<figure class="highlight plaintext"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br></pre></td><td class="code"><pre><span class="line">1: (1)</span><br><span class="line">2: (1) (2)</span><br><span class="line">3: (1)</span><br><span class="line">4: (2) (2)</span><br><span class="line">5: (1)</span><br><span class="line">6: (2)</span><br></pre></td></tr></table></figure>

<p>4 颗技能 1 装饰珠，4 颗技能 2 装饰珠 W1(4)+W2(4)=5+15=20。</p>
<h3 id="运行限制"><a href="#运行限制" class="headerlink" title="运行限制"></a>运行限制</h3><ul>
<li>最大运行时间：1s</li>
<li>最大运行内存: 256M</li>
</ul>
<h1 id="二、思路解析"><a href="#二、思路解析" class="headerlink" title="二、思路解析"></a>二、思路解析</h1><p>毋庸置疑这是一道非常难的动态规划问题，但仔细去分析可以发现这是一道基于0/1背包问题的动态规划题目。</p>
<p>首先我们要有一个动态规划的思路，我进行了以下数理：</p>
<ul>
<li>每件装备有对应的等级，等级最大不超过4，那么我们可以以相同的等级去给这些装饰孔进行统计分类，然后按照相同的等级去遍历这些装饰孔</li>
<li>输入数据中，有m行不同的装饰珠，所以这些装饰珠可以理解为0/1背包问题中的物品，控制外层for循环</li>
<li>这里有一点与0/1背包中不同的地方，就是在0/1背包问题中，对于同样重量的物品，也就是相同的物品，个数对价值的影响是线性的，单个价值不会变，而在这道题目中，个数对于价值所产生的影响是没有规律的，因此要存放到一个数组中，对于不同的数量分别遍历</li>
<li>在外层不同装饰珠的控制下，内层要用装饰珠数量和背包容量相结合的方式（这里又是两层循环）</li>
<li>遍历思路：因为装饰孔中只能镶嵌小于等于该等级的装饰珠，因此从最高等级的装饰孔逐级向下遍历，这样保证只要在最高等级范围内的装饰珠永远是有装入的可能性（以此条件才能构建状态转移方程）</li>
<li>题目中有一个比较坑的地方，就是装饰珠的等级有可能会大于所有的装饰孔，这就是一种无法使用的装饰珠</li>
</ul>
<p>数据结构上：</p>
<p>int dp[a][b] 表示在前a种珠子（包括a）的情况下，装饰孔数量为b时所能产生的最大价值</p>
<p>选取最大值时，比较的是在前a-1种珠子下，与当前加入了a种珠子时对于最大价值的影响，这里还需要一层循环用来控制第a种珠子个数对价值的影响</p>
<p>核心程序：</p>
<figure class="highlight plaintext"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br></pre></td><td class="code"><pre><span class="line">//现将前kind-1种珠子的价值情况存放到当前开放第kind种珠子后的数组中，方便后续处理</span><br><span class="line">for(int j=1;j&lt;=count;j++)</span><br><span class="line">&#123;</span><br><span class="line">	dp[kind][j]=dp[kind-1][j];</span><br><span class="line">&#125;</span><br><span class="line"></span><br><span class="line">//状态转移方程</span><br><span class="line">for (int t = 1; t &lt;= lim[i]; t++)  //对于珠子的每种价值</span><br><span class="line">&#123;</span><br><span class="line">	for (int j = t;j&lt;=count;j++)  //分析背包容量所产生的影响</span><br><span class="line">	&#123;</span><br><span class="line">		dp[kind][j] = max(dp[kind][j], dp[kind - 1][j - t] + value[i][t]);  //比较不放和放两种情况产生的价值</span><br><span class="line">	&#125;</span><br><span class="line">&#125;</span><br></pre></td></tr></table></figure>

<p><strong>另一组测试样例：</strong></p>
<p>输入：</p>
<figure class="highlight plaintext"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br></pre></td><td class="code"><pre><span class="line">5 1 1 1 1 1</span><br><span class="line">9 1 1 1 1 1 1 1 1 1</span><br><span class="line">7 1 1 1 1 1 1 1</span><br><span class="line">3 1 1 1</span><br><span class="line">7 1 1 1 1 1 1 1</span><br><span class="line">9 1 1 1 1 1 1 1 1 1</span><br><span class="line">15</span><br><span class="line">4 3 48 86 129</span><br><span class="line">3 1 3</span><br><span class="line">1 7 45 84 109 154 175 206 233</span><br><span class="line">1 2 35 42</span><br><span class="line">1 6 7 52 67 116 131 156</span><br><span class="line">1 1 32</span><br><span class="line">3 4 12 15 61 82</span><br><span class="line">1 7 5 36 85 88 124 139 163</span><br><span class="line">1 7 29 32 69 101 150 197 210</span><br><span class="line">1 6 9 50 73 102 141 158</span><br><span class="line">3 2 9 44</span><br><span class="line">1 3 1 39 86</span><br><span class="line">1 2 35 78</span><br><span class="line">1 7 3 50 53 96 107 140 175</span><br><span class="line">3 6 16 63 74 89 92 140</span><br></pre></td></tr></table></figure>

<p>输出：</p>
<figure class="highlight plaintext"><table><tr><td class="gutter"><pre><span class="line">1</span><br></pre></td><td class="code"><pre><span class="line">1195</span><br></pre></td></tr></table></figure>

<h1 id="三、AC代码"><a href="#三、AC代码" class="headerlink" title="三、AC代码"></a>三、AC代码</h1><figure class="highlight c"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br><span class="line">24</span><br><span class="line">25</span><br><span class="line">26</span><br><span class="line">27</span><br><span class="line">28</span><br><span class="line">29</span><br><span class="line">30</span><br><span class="line">31</span><br><span class="line">32</span><br><span class="line">33</span><br><span class="line">34</span><br><span class="line">35</span><br><span class="line">36</span><br><span class="line">37</span><br><span class="line">38</span><br><span class="line">39</span><br><span class="line">40</span><br><span class="line">41</span><br><span class="line">42</span><br><span class="line">43</span><br><span class="line">44</span><br><span class="line">45</span><br><span class="line">46</span><br><span class="line">47</span><br><span class="line">48</span><br><span class="line">49</span><br><span class="line">50</span><br><span class="line">51</span><br><span class="line">52</span><br><span class="line">53</span><br><span class="line">54</span><br><span class="line">55</span><br><span class="line">56</span><br><span class="line">57</span><br><span class="line">58</span><br><span class="line">59</span><br><span class="line">60</span><br><span class="line">61</span><br><span class="line">62</span><br><span class="line">63</span><br><span class="line">64</span><br><span class="line">65</span><br><span class="line">66</span><br><span class="line">67</span><br><span class="line">68</span><br><span class="line">69</span><br><span class="line">70</span><br><span class="line">71</span><br><span class="line">72</span><br><span class="line">73</span><br><span class="line">74</span><br><span class="line">75</span><br><span class="line">76</span><br><span class="line">77</span><br><span class="line">78</span><br><span class="line">79</span><br><span class="line">80</span><br><span class="line">81</span><br></pre></td><td class="code"><pre><span class="line"><span class="meta">#<span class="meta-keyword">include</span><span class="meta-string">&lt;bits/stdc++.h&gt;</span></span></span><br><span class="line"><span class="keyword">using</span> <span class="keyword">namespace</span> <span class="built_in">std</span>;</span><br><span class="line"><span class="function"><span class="keyword">int</span> <span class="title">main</span><span class="params">()</span></span></span><br><span class="line"><span class="function"></span>&#123;</span><br><span class="line">    <span class="keyword">int</span> num[<span class="number">7</span>]=&#123;<span class="number">0</span>&#125;;  <span class="comment">//装备对应的装饰孔数量</span></span><br><span class="line">    <span class="keyword">int</span> weapon[<span class="number">7</span>][<span class="number">51</span>]=&#123;<span class="number">0</span>&#125;;  <span class="comment">//装备对应装饰孔的等级</span></span><br><span class="line">    <span class="keyword">int</span> count_grade[<span class="number">5</span>]=&#123;<span class="number">0</span>&#125;;  <span class="comment">//统计相同等级下的个数</span></span><br><span class="line">    <span class="keyword">int</span> grade[<span class="number">10005</span>]=&#123;<span class="number">0</span>&#125;;  <span class="comment">//装饰珠等级</span></span><br><span class="line">    <span class="keyword">int</span> lim[<span class="number">10005</span>]=&#123;<span class="number">0</span>&#125;;  <span class="comment">//装饰珠价值上限个数</span></span><br><span class="line">    <span class="keyword">int</span> value[<span class="number">10005</span>][<span class="number">8</span>]=&#123;<span class="number">0</span>&#125;;  <span class="comment">//装饰珠的价值</span></span><br><span class="line">    <span class="keyword">int</span> m;  <span class="comment">//装饰珠的种类</span></span><br><span class="line">    <span class="keyword">int</span> total=<span class="number">0</span>;</span><br><span class="line">    <span class="keyword">int</span> dp[<span class="number">10001</span>][<span class="number">300</span>]=&#123;<span class="number">0</span>&#125;;  <span class="comment">//dp背包</span></span><br><span class="line">    <span class="keyword">for</span>(<span class="keyword">int</span> i=<span class="number">1</span>;i&lt;=<span class="number">6</span>;i++)</span><br><span class="line">    &#123;</span><br><span class="line">        <span class="built_in">cin</span>&gt;&gt;num[i];</span><br><span class="line">        <span class="keyword">for</span>(<span class="keyword">int</span> j=<span class="number">1</span>;j&lt;=num[i];j++)</span><br><span class="line">        &#123;</span><br><span class="line">            <span class="built_in">cin</span>&gt;&gt;weapon[i][j];  <span class="comment">//等级</span></span><br><span class="line">            count_grade[weapon[i][j]]++;</span><br><span class="line">            total++;</span><br><span class="line">        &#125;</span><br><span class="line">    &#125;</span><br><span class="line">    <span class="built_in">cin</span>&gt;&gt;m;</span><br><span class="line">    <span class="keyword">for</span>(<span class="keyword">int</span> i=<span class="number">1</span>;i&lt;=m;i++)</span><br><span class="line">    &#123;</span><br><span class="line">        <span class="built_in">cin</span>&gt;&gt;grade[i];</span><br><span class="line">        <span class="built_in">cin</span>&gt;&gt;lim[i];</span><br><span class="line">        <span class="keyword">for</span>(<span class="keyword">int</span> j=<span class="number">1</span>;j&lt;=lim[i];j++)</span><br><span class="line">        &#123;</span><br><span class="line">            <span class="built_in">cin</span>&gt;&gt;value[i][j];  <span class="comment">//在第i种珠子下，个数为j所对应的价值</span></span><br><span class="line">        &#125;</span><br><span class="line">    &#125;</span><br><span class="line">    <span class="keyword">int</span> count=<span class="number">0</span>;   <span class="comment">//截止目前开放的孔数</span></span><br><span class="line">    <span class="keyword">int</span> kind=<span class="number">0</span>;  <span class="comment">//已经遍历的珠子种类</span></span><br><span class="line">    </span><br><span class="line">    <span class="comment">//等级从高等级向低等级遍历</span></span><br><span class="line">    <span class="keyword">for</span>(<span class="keyword">int</span> level=<span class="number">4</span>;level&gt;<span class="number">0</span>;level--)</span><br><span class="line">    &#123;</span><br><span class="line">        <span class="keyword">if</span>(count_grade[level]!=<span class="number">0</span>)</span><br><span class="line">        &#123;</span><br><span class="line">            count+=count_grade[level];  <span class="comment">//孔数累加</span></span><br><span class="line">        &#125;</span><br><span class="line">        <span class="keyword">if</span>(count&gt;<span class="number">0</span>)</span><br><span class="line">        &#123;</span><br><span class="line">            <span class="keyword">for</span>(<span class="keyword">int</span> i=<span class="number">1</span>;i&lt;=m;i++)  <span class="comment">//遍历m种珠子</span></span><br><span class="line">            &#123;            </span><br><span class="line">                <span class="keyword">if</span>(grade[i]==level)  <span class="comment">//在该等级下发现新的珠子</span></span><br><span class="line">                &#123;</span><br><span class="line">    </span><br><span class="line">                    kind++;</span><br><span class="line">                    <span class="comment">//现将前kind-1种珠子的价值情况存放到当前开放第kind种珠子后的数组中，方便后续处理</span></span><br><span class="line">                    <span class="keyword">for</span>(<span class="keyword">int</span> j=<span class="number">1</span>;j&lt;=count;j++)</span><br><span class="line">                    &#123;</span><br><span class="line">                        dp[kind][j]=dp[kind<span class="number">-1</span>][j];</span><br><span class="line">                    &#125;</span><br><span class="line">                    </span><br><span class="line">                    <span class="comment">//状态转移方程</span></span><br><span class="line">                    <span class="keyword">for</span> (<span class="keyword">int</span> t = <span class="number">1</span>; t &lt;= lim[i]; t++)  <span class="comment">//对于珠子的每种价值</span></span><br><span class="line">                    &#123;</span><br><span class="line">                        <span class="keyword">for</span> (<span class="keyword">int</span> j = t;j&lt;=count;j++)  <span class="comment">//分析背包容量所产生的影响</span></span><br><span class="line">                        &#123;</span><br><span class="line">                            dp[kind][j] = max(dp[kind][j], dp[kind - <span class="number">1</span>][j - t] + value[i][t]);  <span class="comment">//比较不放和放两种情况产生的价值</span></span><br><span class="line">                        &#125;</span><br><span class="line">                    &#125;</span><br><span class="line">                &#125;</span><br><span class="line">            &#125;</span><br><span class="line">        &#125;</span><br><span class="line">    &#125;</span><br><span class="line">    </span><br><span class="line">    <span class="keyword">int</span> temp=<span class="number">0</span>;</span><br><span class="line">    <span class="keyword">for</span>(<span class="keyword">int</span> i=<span class="number">1</span>;i&lt;=total;i++)</span><br><span class="line">    &#123;</span><br><span class="line">        <span class="keyword">if</span>(temp&lt;dp[kind][i])</span><br><span class="line">        &#123;</span><br><span class="line">            temp=dp[kind][i];</span><br><span class="line">        &#125;</span><br><span class="line">    &#125;</span><br><span class="line">    <span class="built_in">cout</span>&lt;&lt;temp&lt;&lt;<span class="built_in">endl</span>;</span><br><span class="line">    <span class="keyword">return</span> <span class="number">0</span>;</span><br><span class="line">&#125;</span><br></pre></td></tr></table></figure>

<p>参考大佬博客：<a target="_blank" rel="noopener" href="https://blog.csdn.net/weixin_46270935/article/details/115093781">蓝桥杯真题——装饰珠_Hey XIN的博客-CSDN博客_蓝桥杯装饰珠</a> </p>
<p>感谢阅读！也欢迎大家关注小白博主，多多鼓励一下！</p>

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